Question

Show that the gradient of the secant line to the curve `y=x^2+3x` at the points on the curve where `x=3` and `x=3+h` is `h+9`?

Answer 1

We have:

`y=x^2+3x`

When `x=3` we have:

`y =3^2+3(3)`
`\ \ =9+9`
`\ \ =18`

When `x=3+h` we have:

`y = (3+h)^2+3(3+h)`
`\ \ = 9+6h+h^2 + 9 + 3h`
`\ \ = h^2+9h+18`

So the gradient of the secant line is given by:

`m_(sec) = (Delta y)/(Delta x)`

`" " = (y_(3+h) - y_h)/((3+h)-(3))`

`" " = ((h^2+9h+18) - (18))/(3+h-3)`

`" " = (h^2+9h)/(h)`

`" " = h+9 \ \ \` QED

Conclusion

Note that as `h rarr 0` then the secant line becomes the tangent, so in the limit we have:

`m_(tan) = lim_(h rarr 0) m_(sec)`

`" " = lim_(h rarr 0) (h+9)`

`" " = 9`

It should be clear that this final result comes as a direct result of the definition of the derivative. We can also use our knowledge of Calculus to validate this, as:

`y=x^2+3x => dy/dx = 2x + 3`

And so, when `x=3`, we have:

`[dy/dx]_(x=3) = 2(3) + 3 = 9`